Proof: Reduction Example

Let's prove the following theorem:

if not (b = 0), then ((a / b) ⋅ c) ⋅ b = a ⋅ c

First we show that:

(a / b) ⋅ c ⋅ b = a ⋅ (1 / b) ⋅ c ⋅ b (step 3)

Then we use the Reordering theorem to move the terms around:

a ⋅ (1 / b)) ⋅ c ⋅ b = (1 / b) ⋅ b ⋅ a ⋅ c (step 4)

Then we show that:

(1 / b) ⋅ b ⋅ a ⋅ c = 1 ⋅ a ⋅ c (step 7)

Then we simplify the right side as follows:

1 ⋅ a ⋅ c = a ⋅ c (step 9)

Finally, we use the Transitive Property a few times to reach our conclusion.

Proof:

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Given
1 not (b = 0)
Proof Table
# Claim Reason
1 a / b = a ⋅ (1 / b) a / b = a ⋅ (1 / b)
2 (a / b) ⋅ c = (a ⋅ (1 / b)) ⋅ c if a / b = a ⋅ (1 / b), then (a / b) ⋅ c = (a ⋅ (1 / b)) ⋅ c
3 ((a / b) ⋅ c) ⋅ b = ((a ⋅ (1 / b)) ⋅ c) ⋅ b if (a / b) ⋅ c = (a ⋅ (1 / b)) ⋅ c, then ((a / b) ⋅ c) ⋅ b = ((a ⋅ (1 / b)) ⋅ c) ⋅ b
4 ((a ⋅ (1 / b)) ⋅ c) ⋅ b = (((1 / b) ⋅ b) ⋅ a) ⋅ c ((a ⋅ (1 / b)) ⋅ c) ⋅ b = (((1 / b) ⋅ b) ⋅ a) ⋅ c
5 (1 / b) ⋅ b = 1 if not (b = 0), then (1 / b) ⋅ b = 1
6 ((1 / b) ⋅ b) ⋅ a = 1 ⋅ a if (1 / b) ⋅ b = 1, then ((1 / b) ⋅ b) ⋅ a = 1 ⋅ a
7 (((1 / b) ⋅ b) ⋅ a) ⋅ c = (1 ⋅ a) ⋅ c if ((1 / b) ⋅ b) ⋅ a = 1 ⋅ a, then (((1 / b) ⋅ b) ⋅ a) ⋅ c = (1 ⋅ a) ⋅ c
8 1 ⋅ a = a 1 ⋅ a = a
9 (1 ⋅ a) ⋅ c = a ⋅ c if 1 ⋅ a = a, then (1 ⋅ a) ⋅ c = a ⋅ c
10 (((1 / b) ⋅ b) ⋅ a) ⋅ c = a ⋅ c if (((1 / b) ⋅ b) ⋅ a) ⋅ c = (1 ⋅ a) ⋅ c and (1 ⋅ a) ⋅ c = a ⋅ c, then (((1 / b) ⋅ b) ⋅ a) ⋅ c = a ⋅ c
11 ((a ⋅ (1 / b)) ⋅ c) ⋅ b = a ⋅ c if ((a ⋅ (1 / b)) ⋅ c) ⋅ b = (((1 / b) ⋅ b) ⋅ a) ⋅ c and (((1 / b) ⋅ b) ⋅ a) ⋅ c = a ⋅ c, then ((a ⋅ (1 / b)) ⋅ c) ⋅ b = a ⋅ c
12 ((a / b) ⋅ c) ⋅ b = a ⋅ c if ((a / b) ⋅ c) ⋅ b = ((a ⋅ (1 / b)) ⋅ c) ⋅ b and ((a ⋅ (1 / b)) ⋅ c) ⋅ b = a ⋅ c, then ((a / b) ⋅ c) ⋅ b = a ⋅ c

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